Another tough 'elementary' question from my tutor

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"There is nothing to summarize"
I thought that it was an international electronic forum. Everybody in the world can share their idea.
But if it's a scientific platform, then the solution must be constructed. Everybody can't have the same background, that's why communication is a matter, you must make more effort to clarify yourself, and make sure that you understand the other, you should have emphaty.
You can't said :"Your wrong", "that's wrong" without giving further explaination. That simply kills the dialog.
The synthesis that I recommend, allows to go further in the dialog, there won't be the same ideas repeated by different persons.
Otherwise it will just be a dumb to dumb dialog.
 

jasmin_123 said:
To try and clear the mix-up of my R=0 (singularity) and your R->0,
let me ask you about the phase at f=1/(2piRC) of the ideal [R=1/(2piC), not R->1/(2piC)] Wien-bridge bandpass. Is it pi/2, -pi/2, or zero?
Click to expand...

Jasmine,

Are you speaking about the RCseries-RCparallel net, or about the compete bridge (with the 2-R arm)?

Regards

Z
 

Yes, Zorro, I mean the entire bridge and the phase of its differential output, which starts at pi and reaches –pi.
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For the ideal bridge, does the phase go through zero or not?

Added after 23 minutes:

Hi, Mr.MEB,

I hoped that I was very clear:
jasmin_123 said:
There is nothing to summarize: if you do assume zero losses, then steady state means oscillations.
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If you do not like to assume zero losses, this problem is not for you.
Click to expand...

The problem in its original form implies zero losses. If somebody is not ready to accept this condition then why try and solve the problem?

I did not ask whether losses can be zero or not, I just asked what would happen if losses were zero. That's all.

I was not interested in the discussion about losses; I was just interested in a purely theoretical steady state for a purely theoretical case of zero losses.

What else should I say to justify myself? :)

Jasmine
 




Moving electron make always electromagnetic field and store energy on it, in case voltage in cap is equal short time after closing - current is maximal between cap and half of energy i stored as magnetic field around conductors between cap. if current is zero in small time later again, no energy in electromagnetic field exist (drained in work to charge cap from half to full) and ether cap is full charged with full energy and other is emty, current begin go back again and... so on.

So anwer in case 'no loss allowed', 'steady state' is a cap charges oscillating in infinite time between cap, oscillating frequency depend of capacitance and inductance in conductors between cap.

piece-oscillator using this methode in most MCU, PIC and clock source for digital equipment. 180 degree amplifiers in oscillator (simple inverter) only exist to compensate loss in inductance and capacitance (inverter work as negative resistance) in resonant circurit (22 pF cap each side of crystal to ground and crystal itself working as heavy inductance - simulate question above very close)

If inductance and electromagnetic field not allowed in question - question have no coupling to physic and electronic world and fit better laywner, priest and politican people - this guys manipulate always truth...
 

Shouldn't it be this way:

W=1/2*C*U^2=1/2*(2*C)*Ux^2

Ux=U/sqrt(2)

This is the steady state.

At least from the energetic point of view. However a capacitor can not be charged with constant voltage as an inductor can. Constant voltage applied to capacitor results in infinite current. And that is expensive
 

jasmin_123 said:
The steady state in a non-lossless system is obvious: V/2.

I was not interested in the discussion about losses; I was just interested in a purely theoretical steady state for a purely theoretical case of zero losses.
Click to expand...


Even though you are not interested in losses but you assumed that V/2 is the steady state of a system with loss, therefore if you do not account for the loss and add that to the system how would you be able to find the steady state of the lossless system??

I suspect that you came to the calculation of V/2 as the steady state of a system with loss based on the formula Q=VC??(or if it's anything else, please clarify).

Often and usually, loss needs to be calculated and accounted for based on individual conditions (ie. Resistance in each case differ), formulas does not give you the answer with loss included.

There is no part in this formula either Q, C or V that INCLUDE or REPRESENTS LOSS unless you've worked that out and subs into either Q or V.(Which you have not done)

Therefore, you can not use this formula and assume that it gives you the result which includes loss in the system = V/2, because you ignored the loss in the system.

Q=VC is used to calculate for lossless results, it also describe the relationship between Q, V and C in lossless conditions. If oscillation is a behaviour/characteristic of this "lossless component", you would find alternating wave functions such as sine or cosine etc........... as part of the formula because it describe the behaviour of the component in lossless conditions.

Since Q=VC represents lossless result and it does not include alternating wave function, this means oscillation is not part of the steady state in this lossless system.
 

hi iggyboy,
what u said is correct when there is losses... But we cannot transfer charge without spending the energy, which will be accounted as losses. So the energy before combining capacitors and after combining is not same as losses are also there.

U know why losses happen, eventhough the resistance is not there???
It is because of electic filed formed by moved chrage to other capacitor from one capacitor, which rpelles the electrons. As long as voltage differenc is there , there will be flow of electons... so the energy loss is inevitable....

Hope u got what I am trying to say....

If u have anyother openion pls let me know....
 

sivakumar_tumma said:
hi iggyboy,
what u said is correct when there is losses... But we cannot transfer charge without spending the energy, which ...
Click to expand...

Did you mean:

what u said is correct when there is NO losses... But we cannot transfer charge without spending the energy, which ... ?

I do agree with your post except for your statement that the state of charge in the capacitor being charged affects the losses. Why? Electrons are never slowed done by the field in the capacitor, so why should there be losses related with that.
 


when there is losses in the systems energy equating is not correct.

I thought because of electro magnetic forces, energy loss will be there..
 

A nice ques. Does jasmin_123 knows the answer ??? or she also does not know ??

and do you find such ques. in a certain book?? what is it??

thanks

bye
 

The problem in its original form implies zero losses.

I did not ask whether losses can be zero or not, I just asked what would happen if losses were zero. That's all.

I was not interested in the discussion about losses; I was just interested in a purely theoretical steady state for a purely theoretical case of zero losses.

The only right answer in this case are oscillations in the steady state.

Jasmine
 

hi all,

jasmin_123 said:
I did not ask whether losses can be zero or not, I just asked what would happen if losses were zero.
Click to expand...

i believe asking this is like asking "what would happen to the world if we assume 5=3 ?" I think that ideal cases are... ideal. But impossible cases are just... impossible...

unless...

we ask Mr. Schröndiger's cat, which might tell us that both capacitors are charged with the same charge and voltage at the same time, no oscillations, no energy losses, just two states living in the same time and space...

or it might just be too late and i should be going to bed...

good night!
 

Hi,

The energy related explanation can be seen this way. 1/2CV^2 is the energy stored in a capacitor when it charges from 0, because
I = C(dV/dt)
and the energy stored in the capacitor is

∞
W = ∫ IVdt
0
∞
W =∫CVdV
0

W = 1/2C(V^2 - V0^2)
where V0 is the initial voltage across the capacitor. Therefore in the second capacitor(initially discharged) the energy stored at steady state is 1/2C(V/2)^2 and in the first one (initialy charged to V) the energy stored is ( 1/2C(V0^2 - V^2). Therefore the total energy in the combination is unchanged.

∫
 

well... i'll save time to jasmine:

guys, read the previous posts... read the FIRST post, indeed, and you'll understand what jasmine is trying to show is an absolutely ideal scenario, where no resistance, no energy loss, no charge loss (which actually is lost in what subharpe tried to explain)...

my previous post was just aimed for a little laugh...

but seriously, i think people shouldn't even bother to answer if they won't read previous post from a thread...
 

Jasmine, I liked your question actually you remind me myself a few years ago. When I was student(not EE) and I had many questions that none of my professors could answer and sometimes after many weeks I could come up with a satisfactory answer and sometimes not even after years. Try to keep this, I think it is a virtue.


Well, As far as I understood your question(I have not read all the posts), It is simpler instead of considering 2 different capacitors, suppose surface of plates of a capacitor in zero time doubles. Now what actually happens is electrons that had been concentrated on one half of the capacitor now see their playground has doubled. So they madly run to the empty half(driven by potential difference) as they are running the first half of plates starts depletion and some electrons find themselves if they remain in the first half of plate they will have larger playground than the second half that is being swarmed by electrons so they remain there and distribute equally.
But being more scientific it is all about energy and stability. In metal plates most external layer electrons of atoms are in a loose layer and can freely move between their atom and neighboring atoms but if they leave their atom a positive charge remains that should be either filled with themselves(back to their first place) or by neighbor electrons that their move also creates another positive charge that should be filled by an electron and so on for all atoms of the plate. when you have a charged plate and suddenly make the surface double(in time of zero) atoms that have extra negative charge are so willing to distribute it equally so they all the atoms can reach most stable state.
if more than half of electrons go to the second state it leaves more unneutralized charge in the first half than the second half and this needs an extra amount of energy which dose not exist in your system.
 

It is trick question by ommiting inductance of super conductor. Electrons do not move all at once but circuit becomes LC oscillator. If there is no inductance, then we are not talking about electrons. We are talking about magical charge carriers that jump instantly from one place to another or magical capacitors that ocupy same space and appear and dissapear instantly.
Electron interacts with other particles. If there are no other particles it would not matter as everything is relative to something else.


X²-X²=X²-X²
X*(X-X)=(X+X)*(X-X) `/(X-X)
X=X+X `/X
1=2
 

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